# Solving Systems of Linear Equations using the Personal Algebra Tutor ®

### Solving Systems of Linear Equations using the Personal Algebra Tutor illustrates the versatility of the program. The following example will be solved using five different methods.

The Personal Algebra Tutor:- Solves Systems of 2 or 3 Linear Equations Step-by-step
- Covers methods used from beginning algebra thru preCalculus
- Solutions of the following System of Linear Equations produced by the Personal Algebra Tutor will be shown.
Y + X = 5 Equation # 1 Y - 2X = - 1 Equation # 2

- Elementary Algebraic Operations
- Graphing Solution
- Determinants
- Matrices Using Gauss Elimination and Back Substitution
- Matrices Using Gauss-Jordan Elimination

- All methods except the Graphing method can be used for systems of three Linear Equations. An output of the Personal Algebra Tutor is shown below for a System of three Linear Equations using Gauss-Jordan Elimination

## Solution Using Elementary Operations

Y + X = 5 Equation # 1 Y - 2X = - 1 Equation # 2 | Next ... | Subtract: Equation #2 - Equation #1 to eliminate Y 3X = 6 Eq. #1 and #2 Combined | Next ... | Divide by coefficient of variable to solve for X. X = 2 Solution for X | Next choose one of the original equations and | substitute the value of X into that equation. | ... Choosing Equation # 2 Y - 4 = - 1 Y = 3 Solution for Y | **** Solution Set: X = 2, Y = 3,

## Graphical Solution

Y + X = 5 Equation # 1 Y - 2X = - 1 Equation # 2 Plotting Y+X =5 Plot the X-Intercept(Red): Set y=0; Solve for x: (5,0) Plot the Y-Intercept(Blue): Set x=0; Solve for y: (0,5) Draw the line between the points. Finding Slope-Intercept Equation of Line. Solve for 'Y'. Y= -1 X +5 Slope_Intercept Equation Slope = -1 The coefficient of the X term. Y-Intercept = 5 The constant term. Done. Plotting 2X - Y = 1 Plot the X-Intercept(Red): Set y=0; Solve for x: (0.5,0) Plot the Y-Intercept(Blue): Set x=0; Solve for y: (0,-1) Draw the line between the points. Finding Slope-Intercept Equation of Line. Solve for 'Y'. Y= 2 X -1 Slope_Intercept Equation Slope = 2 The coefficient of the X term. Y-Intercept = -1 The constant term. Done. | The lines intersect at x=2, y=3 | The solution is x=2, y=3 ***** Problem Summary ***** Problem Y+X =5 Equation #1 2X - Y = 1 Equation #2 Solution Solution: x=2, y=3

## Solution Using Determinants

Y+X =5 Eq 1 2X-Y=1 Eq 2 Evaluate the Coefficient Determinant, D | 1 1 | D = | | | 2 -1 | D = 1*(-1) -(1)*(2) = -3 | 5 1 | Dx= | | | 1 -1 | Dx= 5*(-1) -(1)*(1) = -6 Matrix Solution Dx -6 X = ---- = ---- = 2 D -3 | 1 5 | Dy= | | | 2 1 | Dy= 1*(1) -(5)*(2) = -9 Dy -9 Y = ---- = ---- = 3 D -3 **** X = 2, Y = 3 Solution is Consistent.

## Matrix Solution Using Gauss Elimination and Back Substitution

Y+X =5 Eq 1 2X-Y=1 Eq 2 | 1 1 | Mc= | 2 -1 | Coefficient Matrix | 1 1 | 5 | Ma= | 2 -1 | 1 | Augmented Matrix Solve using the Augmented Matrix | 1 1 | 5 | M = | 0 3 | 9 | 2 R1 - R2 -> R2 Force all Row leading non-zero coefficients to 1. | 1 1 | 5 | X +1 Y = 5 M = | 0 1 | 3 | Y = 3 Back Substitution Y = 3 X = 5 -(1)Y = 5 -(1)(3) X = 2 **** X = 2, Y = 3 Solution is Consistent.===========================================

## Matrices Using Gauss-Jordan Elimination

Y+X =5 Eq 1 2X-Y=1 Eq 2 | 1 1 | Mc= | 2 -1 | Coefficient Matrix | 1 1 | 5 | Ma= | 2 -1 | 1 | Augmented Matrix Solve using the Augmented Matrix | 1 1 | 5 | M = | 0 3 | 9 | 2 R1 - R2 -> R2 | -3 0 | -6 | R2 - 3 R1 -> R1 M = | 0 3 | 9 | Force all Row leading non-zero coefficients to 1. | 1 0 | 2 | X = 2 Mj= | 0 1 | 3 | Y = 3 **** X = 2, Y = 3 Solution is Consistent.

## System of 3 Linear Equations Example Matrices Using Gauss-Jordan Elimination

2X-Y+Z=44 Eq 1 -X+3Y-2Z=-53 Eq 2 5X-6Y-Z=19 Eq 3 | 2 -1 1 | Mc= | -1 3 -2 | Coefficient Matrix | 5 -6 -1 | | 2 -1 1 | 44 | Ma= | -1 3 -2 | -53 | Augmented Matrix | 5 -6 -1 | 19 | Solve using the Augmented Matrix | 2 -1 1 | 44 | M = | -1 3 -2 | -53 | | 0 7 7 | 182 | 5 R1 - 2 R3 -> R3 | 2 -1 1 | 44 | M = | 0 5 -3 | -62 | R1 + 2 R2 -> R2 | 0 7 7 | 182 | | 2 -1 1 | 44 | M = | 0 5 -3 | -62 | | 0 0 -56 | -1344 | 7 R2 - 5 R3 -> R3 | -112 56 0 | -1120 | - R3 + 56 R1 -> R1 M = | 0 5 -3 | -62 | | 0 0 -56 | -1344 | | -112 56 0 | -1120 | M = | 0 280 0 | 560 | -3 R3 + 56 R2 -> R2 | 0 0 -56 | -1344 | | -112 56 0 | -1120 | R1/280 -> R1 M = | 0 1 0 | 2 | | 0 0 -56 | -1344 | | 112 0 0 | 1232 | 56 R2 - R1 -> R1 M = | 0 1 0 | 2 | | 0 0 -56 | -1344 | Force all Row leading non-zero coefficients to 1. | 1 0 0 | 11 | X = 11 Mj= | 0 1 0 | 2 | Y = 2 | 0 0 1 | 24 | Z = 24 **** X = 11, Y = 2, Z = 24 Solution is Consistent.